Aka Posted April 30, 2006 Posted April 30, 2006 the folowing reaction proceeds with a 70% yeild C6H6 + HNO3 = C6H5NO2 + H2O calculate the mass of C6H5NO2 expected if 12.8g of C6H5 reacts with excess HNO3 How do you do this problem? and If 3.45 x 10^23 formula units of Hg(NO3)2 are reacted with excess of Na2S, what mass of HgS can be expected if this process occurs with 97% yield? What are fomula units?
hybrid04 Posted May 1, 2006 Posted May 1, 2006 Basically you have to look at the forumulas. Since HNO3 is in excess the amount of product produced is dependant on the other reagent. The mole ratio given from the reaction is 1mol C6H6 produces 1 mol C6H5NO2. Remember it is a mole not a mass ratio. So 1g doesnt make one g. Figure out how many moles of starting material that equals the number of moles of product and then multiply by the molar weight. This gives you 100% yield (or theoretical yield). Multiply that by .71 and you have 71% yield. For the second one im not sure exactly what 'formula units' are but you could still use the same process and get formula units in the end. You need to find the balanced formula for the reaction and then just match the coefficients. 1 for 1 or 1 for 2 etc and then mulitply out the forumla units.
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